Let $A$ denote an algebra finite dimensional, basic, and connected algebra over a algebraically closed field $K$. We denote by $mod A$ the abelian category whose objects are finitely generated right modules over $A$.

Let $T$ a tilting module. Denote the torsion pair induced by $T$ in $mod A$ by $(\mathcal{T}(T),\mathcal{F}(T))$ (http://en.wikipedia.org/wiki/Tilting_theory).

The book "Elements of the Representation Theory of Associative Algebras: Volume 1: Techniques of Representation Theory" (http://books.google.com.br/books/about/Elements_of_the_Representation_Theory_of.html?id=ayNHpi3tYhQC&redir_esc=y) has an exercise that is hard to do:

If $N$ is an object of the $\mathcal{F}(T)$, then $$ pd \; Ext^1_A (T,N) \leq 1 + max(1, pd \; N). $$ Anyone have an idea?

Ps.: $pd$ is the projective dimension.


In this answer I freely use results and notation from the book you refer to. Let me know if I need to fill in more details.

Let $U=\operatorname{Ext}^1_A(T,N)$. We can assume $\operatorname{pd}_B (U)=n>2$ since the statement holds trivially for smaller values. Let $$0 \rightarrow \Omega_B U \rightarrow P_B \rightarrow U \rightarrow 0$$ be the first step in a projective $B$-resolution of $U$.

Since $N \in \mathcal F(T)$, we have $U \otimes_B T=0$ and $\operatorname{Tor}^B_1(U,T) \cong N$ as $A$-modules. Therefore, when we apply $- \otimes_B T$ to the sequence above, we obtain an exact sequence of $A$-modules $$0 \rightarrow N \rightarrow \Omega_B(U) \otimes_B T \rightarrow T' \rightarrow 0,$$ where $T'$ is in $\operatorname{add}T$.

The module $\Omega_B(U)$ is in the category $\mathcal Y(T)$ since it is a submodule of a projective $B$-module. There is an exact equivalence $- \otimes_B T \colon \mathcal Y(T) \rightarrow \mathcal T(T)$, and therefore $$\operatorname{pd}_A (\Omega_B(U) \otimes_B T) \geq \operatorname{pd}_B (\Omega_B(U))=n-1>1.$$ Since $\operatorname{pd}_A (T') \leq 1$, it follows from the exact sequence that $$\operatorname{pd}_A(N)=\operatorname{pd}_A (\Omega_B(U) \otimes_B T).$$ Combining with the previous inequality we get $$\operatorname{pd}_A(N) \geq \operatorname{pd}_B (U)-1$$ which proves the statement.

  • 1
    $\begingroup$ There is also a proof in Ringel's book Tame Algebras and Integral Quadratic Forms, but I don't have access to it at the moment. $\endgroup$ – Dag Oskar Madsen Oct 3 '13 at 20:09

$T$ has projective dimension at most one, so take such a resolution:

$$0 \to P_0\to P_1\to T \to 0$$ and Hom into $N$. Using $Hom(T,N)=0$, one gets: $$0\to Hom(P_1,N) \to Hom(P_0,N) \to Ext^1(T,N)\to 0 $$

The two modules on left have projective dimension $pd(N)$ or is $0$, so standard inequalities for projective dimension in an exact sequence shows that $pd(Ext^1(T,N)) \leq 1 +pd(N)$.

  • $\begingroup$ Thank you Hailong, my problem is that I cannot see what is the $B = End(T)$ module structure of the $k$ vector spaces $Hom_A(P_0, N)$ and $Hom_A(P_1, N)$ left alone to compute the projective dimension. $\endgroup$ – Vasco Oct 2 '13 at 21:05
  • $\begingroup$ I am not sure I understand your concern. They are all $A$-modules, aren't they? $\endgroup$ – Hailong Dao Oct 2 '13 at 21:49
  • $\begingroup$ They are $K$-modules. If $P_0$ and $P_1$ were bimodules over $A$, else $Hom_A (P_0,N)$ and $Hom_A (P_1,N)$ would right modules over $A$. In this book, has a result that shows an inequality similar to this exercise. "Let $A$ be an algebra, $T_A$ be a tilting module, and $B=End T_A$. If $M \in \mathcal{T}(T)$, then $pd Hom_A(T,M) \leq pd M$" (p215). The projective dimension of the $Hom_A(T,M)$ is over module structure of the $B= End(T)$. $\endgroup$ – Vasco Oct 3 '13 at 15:14
  • $\begingroup$ I see, so the first pd is over B, and the second one is over A? $\endgroup$ – Hailong Dao Oct 3 '13 at 20:47
  • $\begingroup$ Exactly. The notation of answer Dag Oskar Madsen is better. $\endgroup$ – Vasco Oct 4 '13 at 22:21

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